What is the extreme f(x)=x^3-6x^2+3 ?

The extreme f(x)=x^3-6x^2+3 is Maximum(0,3),Minimum(4,-29)

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extreme f(x)=x^3-6x^2+3

extreme points

f (x) = x^{3} - 6 x^{2} + 3

Maximum (0, 3), Minimum (4, - 29)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

x^{3} - 6 x^{2} + 3 : 3

Maximum (0, 3)

Plug

x = 4

into

x^{3} - 6 x^{2} + 3 : - 29

Minimum (4, - 29)

Maximum (0, 3), Minimum (4, - 29)

f (x) = x^{2} + 9

\frac{4}{x + 6}

f (x, y) = 113 x + 15 y - x^{2} - 2 y^{2} + 3 x y

f (x) = - x^{3} + 6 x^{2} + 15 x + 4

\frac{csc ( x )}{1 + csc ( x )}

What is the extreme f(x)=x^3-6x^2+3 ?
The extreme f(x)=x^3-6x^2+3 is Maximum(0,3),Minimum(4,-29)